Question

In a study designed to measure various aspects of multicultural sensitivity, simple random samples of 15 male and 15 female clinical psychologists were given several tests. On one of the tests, the two sets of sample scores were: Male Female 63 58 71 62 68 65 60 63 69 62 70 58 6962 72 66 6960 63 61 5860 65 70 65 64 70 67 74 63 Do these scores indicate a difference between men and women with respect to this aspect of multicultural sensitivity? Begin by identifying a statement to be tested, the random variable(s) involved and any assumptions you make about them, level of significance, the statistical hypotheses to be tested, the test statistic and critical region. Then, either by hand or using SPSS (provide all SPSS output you use), do the hypothesis test, make a decision about the hypotheses, and draw conclusions as to whether there are differences based on qender.

Answer 1

The null and alternate hypothesis can be written as:

H₀: There is no significant difference between men and women with respect to aspect of multicultural sensitivity i.e. $\mu_{1} = \mu_{2}$

H₁: There is significant difference between men and women with respect to aspect of multicultural sensitivity i.e. $\mu_{1} \neq \mu_{2}$

X	$X-\overline{X}$	$[X-\overline{X}]^{2}$	Y	$Y-\overline{Y}$	$[Y-\overline{Y}]^{2}$
63	-4.07	16.56	58	-4.73	22.37
71	3.93	15.44	62	-0.73	0.5329
68	0.93	0.865	65	2.27	5.1529
60	-7.07	49.98	63	0.27	0.0729
69	1.93	3.72	62	-0.73	0.5329
70	2.93	8.58	58	-4.73	22.3729
69	1.93	3.72	62	-0.73	0.5329
72	4.93	24.305	66	3.27	10.6929
69	1.93	3.725	60	-2.73	7.4529
63	-4.07	16.565	61	-1.73	2.9929
58	-9.07	82.265	60	-2.73	7.4529
65	-2.07	4.285	70	7.27	52.8529
65	-2.07	4.285	64	1.27	1.6129
70	2.93	8.585	67	4.27	18.2329
74	6.93	48.025	63	0.27	0.0729
X-67.07		290.905	у 62.73		152.9306

$S^{2}= \frac{\sum(X-\overline{X})^{2}+\sum({Y-\overline{Y}})^{2}}{n_{1}+n_{2}-2}$

=
290.905 + 152.9306 15+15- 2 443.8356 28 - 15.8513
(where
67.07. У 62.73
)

Now,

$t = \frac{\overline{x}-\overline{y}}{\frac{S}{\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}}$

S = 3.981

Also,

67.07. У 62.73

67.07-62.73

= 0.3981

Degrees of Freedom =

n1 + n2-2=15+15-2=28

Now, the tabulated value of t for 28 at 0.025 level of significance = 2.048 (see the table below)

Since magnitude of calculated value is much less than the tabulated value, the null hypothesis H₀ is accepted.

Table of the Student's t-distribution The table gives the values of la; v where Pr(Ty) a, with v degrees of freedom 0.005 0.001 0.0005 3.078 6.3142.076 31.821 63.657 318.310 636.620 9.925 22.326 31.598 5.841 10.213 2.924 7.173 8.610 5.893 6.869 2.920 6.965 1.533 2.776 2.571 3.747 3.365 4.604 4.032 2.015 2.447 3.143 2.998 2.896 2.262 2.821 2.228 2.764 5.208 5.959 4.785 5.408 5.041 4.297 4.781 4.144 4.587 2.365 2.306 1.397 1.383 3.355 3.250 4.501 2.201 2.718 3.055 3.012 2.624 2.977 4.025 4.437 3.930 4.318 3.852 4.221 3.787 4.140 602 2.947 3.733 4.073 2.650 1.345 1.341 1.753 2.1312. 1.337 2.921 2.583 2.567 2.552 2.539 2.528 3.686 4.015 3.646 3.965 3.610 3.922 3.579 3.883 3.552 3.850 2.110 1.328 1.325 2.093 2.086 2.878 2.861 2.845 20 2.518 2.508 2.819 3 2.500 2.492 2.485 1.323 1.321 2.080 2.074 2.069 2.064 2.060 2.831 2.807 2.797 2.787 3.527 3.819 3.505 3.792 3.485 3.767 3.467 3.745 3.450 3.725 23 2.056 2.052 2.048 2.045 2.042 2.479 2.473 2.467 2.462 2.457 3.435 3.707 3.421 3.690 3.408 3.674 3.396 3.659 3.385 3.646 26 2.779 28 29 1.303 2.021 2.000 2.423 2.704 2.660 2.617 2.576 2.390 2.358 2.326 3.307 3.551 3.232 3.460 3.160 3.373 3.090 3.291 60 1.289